Tìm x :
a. ( 2x - 1 ) = ( 2x - 1 )-10
b. 3x + 3x + 2 = 210
c. 2x = 4y - 1 và 37y = 3x + 8
Tìm x biết :
a) (3x³ + x² – 13x + 5) : (x² + 2x – 1) = 10
b) (x⁴ – 2x² – 8) : (x – 2) = 0
c) (x²-4x) : (x²-8x+16) = 0
\(a,\Leftrightarrow\dfrac{3x^3+6x^2-3x-5x^2-10x+5}{x^2+2x-1}=10\\ \Leftrightarrow\dfrac{3x\left(x^2+2x-1\right)-5\left(x^2+2x-1\right)}{x^2+2x-1}=10\\ \Leftrightarrow3x-5=10\Leftrightarrow3x=15\Leftrightarrow x=5\\ b,\Leftrightarrow\left(x^4+2x^2-4x^2-8\right):\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x^2-4\right)\left(x^2+2\right)\right]:\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\left(x^2+2\right)\right]:\left(x-2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x^2+2\right)=0\Leftrightarrow x=-2\left(x^2+2>0\right)\\ c,\Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-4\right)^2}=0\Leftrightarrow\dfrac{x}{x-4}=0\Leftrightarrow x=0\)
b: \(\Leftrightarrow x^4-4x^2+2x^2-8=0\)
hay x=-2
Tìm x biết
a) -2x(x+3)+x(2x-1)=10
b) (2/3x)(9x/2+1/4)-(3x2x+2)=3
a: -2x(x+3)+x(2x-1)=10
=>-2x^2-6x+2x^2-x=10
=>-7x=10
=>x=-10/7
b: Sửa đề: 2/3x(9/2x+1/4)-(3x^2+2)=3
=>3x^2+1/6x-3x^2-2=3
=>1/6x-2=3
=>x=30
a) (2x - 1)2 - (2x + 5)(2x + 1) = 10
b) 92 (x - 1) + 25 .(1 - x) = 0
c) x2 + 3x - 4 = 0
`#040911`
`a)`
\(\left(2x-1\right)^2-\left(2x+5\right)\left(2x+1\right)=10\)
\(\Leftrightarrow 4x^2 - 4x + 1 - (4x^2 + 12x + 5) = 10 \\ \Leftrightarrow 4x^2 - 4x + 1 - 4x^2 - 12x - 5 = 10 \\ \Leftrightarrow (4x^2 - 4x^2) - (4x + 12x) + (1 - 5) = 10 \\ \Leftrightarrow -16x - 4 = 10 \Leftrightarrow -16x = 10 + 4 \\ \Leftrightarrow -16x = 14 \\ \Leftrightarrow x = \dfrac{-7}{8}\)
Vậy, `x = -7/8`
`b)`
`9^2(x - 1) + 25(1 - x) = 0`
`<=> 9^2(x - 1) - 25(x - 1) = 0`
`<=> (x - 1)(9^2 - 5^2) = 0`
`<=>`\(\left[{}\begin{matrix}x-1=0\\9^2-5^2=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=1\\56=0\left(\text{vô lý}\right)\end{matrix}\right.\)
Vậy, `x = 1`
`c)`
`x^2+3x - 4 = 0`
`<=> x^2 + 4x - x - 4 = 0`
`<=> (x^2 - x) + (4x - 4) = 0`
`<=> x(x - 1) + 4(x - 1) = 0`
`<=> (x + 4)(x - 1) = 0`
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\\ \text{Vậy, }x\in\left\{-4;1\right\}\)
a: =>4x^2-4x+1-(4x^2+2x+10x+5)=10
=>4x^2-4x+1-10-4x^2-12x-5=0
=>-16x-4=0
=>x=-1/4
b: =>(x-1)(9^2-25)=0
=>x-1=0
=>x=1
c: =>x^2+4x-x-4=0
=>(x+4)(x-1)=0
=>x=1 hoặc x=-4
Các bạn giúp mình nha !!! Thanks các bạn !!!
a) 6xz-7x²/4y²+9xz+7x²/4y²
b) 5x+10/4x-8:2x+4/4-2x
c) (2/2x-1 - 2/1-2x):4x²/4x²-1
d) 5b/b²-9.b-3/10b
e) (5/3(x+2)² - 2x/3x+6 + 1/3).x+2/x-3
f) (x/x-y - y/x+y + 2xy/x²-y²):x+y/2x-2y.
Bài1: Rút gọn các biểu thức sau
a, 3x^2-2x:(5+1,5x)+10
b, 7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)
Bài2 : tìm x
a, 3(2x-1)-5(x-3) + 6(3x-4) =24
b, 2x^2 + 3(x^2-1) = 5x(x+1)
c,2x(5-3x) +2x(3x+5) - 3(x-7) =3
Tìm x:
a) 3x.(1 - 2x) + 6x^2 - 7x = 8.(1 - 2x) - 9
b) 2x.(1 + 3x) - 3x.(4 +2x) = 3x - 4
1.CMR:
a) 3.\(\left(x^2+y^2+z^2\right)-\left(x-y\right)^2\) \(-\left(y-z\right)^2-\left(z-x\right)^2=\left(x+y+z\right)^2\)
1) tính nhanh giá trị biểu thức:
a) x^2 + 4y^2 - 4xy tại x=18; y=4
b) (2x + 1)^2 + (2x - 1)^2 - 2 (1 + 2x) (1 - 2x) tại x = 100
2) tìm x biết :
a) 7x^2 -28 =0 b) 2/3x (x^2 - 4) = 0 c) 2x (3x - 5) - (5 - 3x) = 0
d) (2x - 1)^2 -25 = 0
3) phân tích các đa thức sau thành nhân tử :
a) 2(x - 3) - y (x - 3) b) x^3 + 3x^2 - 3x - 1 c) x^2 + 5xy d) x^2 - x - y^2 -y
e) x^2 - 9y^2 +2x +1 f) x^2 - 2x - 4y^2 - 4y g) 10x +15y h) x^2 - 2xy + y^2 - 4
i) 4x - 4y + x^2 - 2xy + y^2 k) x^4 - 4x^3 - 8x^2 - 8x l) x^3 + x^2 - 4x - 4
n) x^3 + x^2y - xy^2 - y^3 o) x^2 - y^2 - 2x - 2y p) x^2 - y^2 - 2x + 2y
q) 2x + 2y - x^2 - xy r) x^2 - 25 + y^2 + 2xy s) x^3 - 2x^2 + x
t) 12x^2y - 18xy^2 - 30y^2 u) 36 - 12x + x^2 v) 3x^2 - 3xy
Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Bài 1: Tìm x € Z a)1−3x chia hết cho x−2 b)3x+2 chia hết cho 2x+1 Bài 2: Tìm các số nguyên a)x(3−y)−y=0 b)xy+2x+2y=0 c)xy−2x+4y=1 d)x(y+1)+y=0
Bài 1:a) Ta có: \(1-3x⋮x-2\)
\(\Leftrightarrow-3x+1⋮x-2\)
\(\Leftrightarrow-3x+6-5⋮x-2\)
mà \(-3x+6⋮x-2\)
nên \(-5⋮x-2\)
\(\Leftrightarrow x-2\inƯ\left(-5\right)\)
\(\Leftrightarrow x-2\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{3;1;7;-3\right\}\)
Vậy: \(x\in\left\{3;1;7;-3\right\}\)
b) Ta có: \(3x+2⋮2x+1\)
\(\Leftrightarrow2\left(3x+2\right)⋮2x+1\)
\(\Leftrightarrow6x+4⋮2x+1\)
\(\Leftrightarrow6x+3+1⋮2x+1\)
mà \(6x+3⋮2x+1\)
nên \(1⋮2x+1\)
\(\Leftrightarrow2x+1\inƯ\left(1\right)\)
\(\Leftrightarrow2x+1\in\left\{1;-1\right\}\)
\(\Leftrightarrow2x\in\left\{0;-2\right\}\)
hay \(x\in\left\{0;-1\right\}\)
Vậy: \(x\in\left\{0;-1\right\}\)
Bài 1 :
a, Có : \(1-3x⋮x-2\)
\(\Rightarrow-3x+6-5⋮x-2\)
\(\Rightarrow-3\left(x-2\right)-5⋮x-2\)
- Thấy -3 ( x - 2 ) chia hết cho x - 2
\(\Rightarrow-5⋮x-2\)
- Để thỏa mãn yc đề bài thì : \(x-2\inƯ_{\left(-5\right)}\)
\(\Leftrightarrow x-2\in\left\{1;-1;5;-5\right\}\)
\(\Leftrightarrow x\in\left\{3;1;7;-3\right\}\)
Vậy ...
b, Có : \(3x+2⋮2x+1\)
\(\Leftrightarrow3x+1,5+0,5⋮2x+1\)
\(\Leftrightarrow1,5\left(2x+1\right)+0,5⋮2x+1\)
- Thấy 1,5 ( 2x +1 ) chia hết cho 2x+1
\(\Rightarrow1⋮2x+1\)
- Để thỏa mãn yc đề bài thì : \(2x+1\inƯ_{\left(1\right)}\)
\(\Leftrightarrow2x+1\in\left\{1;-1\right\}\)
\(\Leftrightarrow x\in\left\{0;-1\right\}\)
Vậy ...
Tìm x biết:
a) (3x³ + x² – 13x + 5) : (x² + 2x – 1) = 10
b) (x⁴ – 2x² – 8) : (x – 2) = 0
c) \(\dfrac{x^2-4x}{x^2-8x+16}\)= 0
b: \(\Leftrightarrow x^4-4x^2+2x^2-8=0\)
\(\Leftrightarrow x+2=0\)
hay x=-2
Bài 1: Phân tích đa thức sau :
a)2x(xy+y^2-3)
b)(x-y)(2x+y)
c)(x-2y)^2
d)(2x-y)(y+2x)
bài 2: Phân tích các đơn thức thành nhân tử
a)3x^2-3xy
b)x^2-4y^2
c)3x-3y+xy-y^2
d)x^2-1+2y-y^2
Bài 3: Tìm x biết:
a)3x^2-6x=0
b)Tìm x,y thuộc z biết: x^2+4y^2-2xy=4
Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)